$$
\begin{eqnarray*}
a_1 & = & \int_{0}^{\theta}\sin^{2}{x}\tan{x}\,dx \\
& = & \int_{0}^{\theta}(1-\cos^2{x})\tan{x}\,dx \\
& = & \int_{0}^{\theta}(\tan{x}-\sin{x}\cos{x})\,dx \\
& = & \int_{0}^{\theta}\left(\tan{x}-\frac{1}{2}\sin{2x}\right)\,dx \\
& = & \biggl[-\log{\bigl|\cos{x}\bigr|} +\frac{1}{4}\cos{2x} \biggr]_{0}^{\theta}\\
& = &-\log{(\cos{\theta})}-\frac{1}{4}(1-\cos{2\theta}) \\
& = &-\log{(\cos{\theta})}-\frac{1}{2}\sin^2{\theta} .
\end{eqnarray*}
$$

$0 \leqq t <\pi/2$ において関数 $\sin{t}$ 及び $\tan{t}$ は単調に増加するから， $0 \leqq x \leqq \theta < \pi/2$ に注意すれば
\[
0 \leqq \sin^{2n}{x}\tan{x} \leqq \sin^{2n}{\theta}\tan{\theta}
\]
が成立する．したがって
\[
0 \leqq a_n = \int_{0}^{\theta} \sin^{2n}{x}\tan{x}\,dx \leqq \int_{0}^{\theta} \sin^{2n}{\theta}\tan{\theta}\,dx = \theta \sin^{2n}{\theta}\tan{\theta}
\]
が成り立つ．ここで，$0 \leqq \sin{\theta} < 1$ より
\[
\lim_{n \to \infty}\sin^{2n}{\theta} = 0
\]
であるから，はさみうちの原理より
\[
\lim_{n \to \infty}a_n = 0.
\]

$$
\begin{eqnarray*}
a_{n+1} & = & \int_{0}^{\theta} \sin^{2n+2}{x}\tan{x}\,dx \\
& = & \int_{0}^{\theta} (1-\cos^2{x})\sin^{2n}{x}\tan{x}\,dx \\
& = & \int_{0}^{\theta} \left(\sin^{2n}{x}\tan{x}-\cos^2{x}\sin^{2n}{x}\tan{x}\right)\,dx \\
& = & \int_{0}^{\theta}\sin^{2n}{x}\tan{x}\,dx-\int_{0}^{\theta}\cos{x}\sin^{2n+1}{x}\,dx \\
& = & a_n-\biggl[\frac{1}{2n+2}\sin^{2n+2}{x} \biggr]_{0}^{\theta}\\
& = & a_n-\frac{\sin^{2n+2}{\theta}}{2n+2}
\end{eqnarray*}
$$
が成り立つ．したがって
\[
a_{n+1}-a_n =-\frac{\sin^{2n+2}{\theta}}{2n+2}
\]
である．

先の結果から

$$
\begin{eqnarray*}
\sum_{n=1}^{N}\frac{\sin^{2n+2}{\theta}}{2n+2} & = & \sum_{n=1}^{N}(a_n-a_{n+1}) \\
& = & (a_1-a_2) + (a_2-a_3) + \cdots + (a_{N-1}-a_{N}) + (a_{N}-a_{N+1}) \\
& = & a_1-a_{N+1} \\
& = & -\log{(\cos{\theta})}-\frac{1}{2}\sin^2{\theta}-a_{N+1}
\end{eqnarray*}
$$
が成り立つ．ここで
\[
\lim_{N \to \infty}a_{N+1} = 0
\]
であるから，$N \to \infty$ の極限を考えて
\[
\sum_{n=1}^{\infty}\frac{\sin^{2n+2}{\theta}}{2n+2} =-\log{(\cos{\theta})}-\frac{1}{2}\sin^2{\theta}
\]
を得る．さらに式を変形すると
$$
\begin{eqnarray*}
\sum_{n=1}^{\infty}\frac{\sin^{2n+2}{\theta}}{2n+2} & = &-\log{(\cos{\theta})}-\frac{1}{2}\sin^2{\theta} \\
\sum_{n=1}^{\infty}\frac{\sin^{2n+2}{\theta}}{n+1} & = & -2\log{(\cos{\theta})}-\sin^2{\theta} \\
\sum_{n=2}^{\infty}\frac{\sin^{2n}{\theta}}{n} & = &-\log{(\cos^2{\theta})}-\sin^2{\theta} \\
\frac{\sin^2{\theta}}{1} + \sum_{n=2}^{\infty}\frac{\sin^{2n}{\theta}}{n} & = & \log{\left(\frac{1}{\cos^2{\theta}}\right)} \\
\sum_{n=1}^{\infty}\frac{\sin^{2n}{\theta}}{n} & = & \log{(1 + \tan^2{\theta})}
\end{eqnarray*}
$$
を得る．

$\theta \neq 0$ のとき，$\sin^2{\theta} \neq 0$ であることに注意して，$\displaystyle a = \frac{1}{\sin^2{\theta}} ~~ ( > 1)$ とおく．このとき
\[
1 + \tan^2{\theta} = \frac{1}{\cos^2{\theta}} = \frac{1}{1-\sin^2{\theta}} = \frac{1}{1-(1/a)} = \frac{a}{a-1}
\]
が成り立つから，級数 $(\heartsuit)$ の計算結果にこれを代入すれば

\[
\sum_{n=1}^{\infty}\frac{1}{n \cdot a^n} = \log{\frac{a}{a -1}}\quad (a > 1)
\]
を得る．

\[
\sum_{n=1}^{\infty}\frac{1}{n\cdot 2^n} = \log{2} \qquad \left(\theta = \frac{\pi}{4}\right)
\]
\[
\sum_{n=1}^{\infty}\frac{1}{n\cdot 4^n} = \log{\frac{4}{3}} \qquad \left(\theta = \frac{\pi}{6}\right)
\]
\[
\sum_{n=1}^{\infty}\frac{3^n}{n \cdot 4^n} = 2\log{2} \qquad \left(\theta = \frac{\pi}{3}\right)
\]

\[
\sum_{n=1}^{\infty}\frac{(2-\sqrt{3})^n}{n \cdot 4^n} = \log{(8-4\sqrt{3})} \qquad \left(\theta = \frac{\pi}{12}\right)
\]
\[
\sum_{n=1}^{\infty}\frac{(3-\sqrt{5})^n}{n \cdot 8^n} = \log{\left(\frac{10-2\sqrt{5}}{5}\right)} \qquad \left(\theta = \frac{\pi}{10}\right)
\]
\[
\sum_{n=1}^{\infty}\frac{(2-\sqrt{2})^n}{n \cdot 4^n} = \log{(3-2\sqrt{2})} \qquad \left(\theta = \frac{\pi}{8}\right)
\]
\[
\sum_{n=1}^{\infty}\frac{(5-\sqrt{5})^n}{n \cdot 8^n} = \log{(6-2\sqrt{5})} \qquad \left(\theta = \frac{\pi}{5}\right)
\]